Q: If $f$ is Riemann integrable over $[0,1]$, then there exists a differentiable function $F$ on $[a, b]$ such that $F'=f$ and $\int_a^b f(x)dx = F(b)-F(a)$.

Here's what I'm thinking so far:

If $\alpha = x$, then you can say that $f$ is also Riemann-Stieltjes integrable on $[0,1]$.

Then, since $f$ must be a bounded real-valued function due to being Riemann integrable, we can use the theorem that states "For $0\le x \le 1$, put $F(x) = \int_0^x f(t) dt$. Then $f$ is continuous on $[0,1]$

In order for $F$ to be differentiable, $f$ must be continuous on $[0,1]$, so we add that as an assumption

In order for $F$ to be differentiable, $F$ should be continuous, not $f$. Otherwise, Riemann integrable would imply continuous (by your argument), and it clearly doesn't.
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JavierDec 5 '12 at 13:19

I agree that Riemann integrable does not imply continuous, but I'm having trouble making this leap. It seems like there will have to be some other requirement on f in order to guarantee such an F exists. I guess I'm thinking f needs to be continuous since its derivative needs to be F.
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GradStudentDec 5 '12 at 13:22

What are you trying to do? The answer to the question is "no" (take any function with a jump discontinuity). Do you wish to find additional conditions on $f$ that would give an affirmative answer to the modified question?
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David MitraDec 5 '12 at 13:30

I am trying to either prove the statement, or give a counterexample. I am thinking I need a counterexample at this point, but am having trouble writing one.
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GradStudentDec 5 '12 at 13:36

3 Answers
3

The question is clearly false. Suppose it were true, and let $f = x$ on $[0,1]$. Let $g = f$ on $[0,1)$ and $g(1) = 0$. Then we have that $\int_a^bf(x) dx = \int_a^b g(x) dx$, since the functions differ at only a finite number of points. So your requirement would necessitate a differentiable function $F$ satisfying $0 = F'(1) = 1$.

Your question has a negative answer. For instance the function defined by $f(x)=\cases{-1,& $-1\le x<0$ \cr 1,&$ 0\le x\le 1$}$ is Riemann integrable over $[-1,1]$, but is not the derivative of any function (since derivatives have the intermediate value property as a result of Darboux's Theorem).

The Darboux Theorem you mentioned is not something we have covered in class, so I am not sure how else to show that f(x) is not the derivative of something.
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GradStudentDec 5 '12 at 16:25

@GradStudent: Darboux's Theorem is the underlying conceptual reason behind counterexamples like this, but it's certainly not needed to check the lack of differentiability in this case. The antiderivative of a piecewise constant jump function like this is a piecewise linear function with a corner point. The left and right hand derivatives at this point are different, so the function is not differentiable.
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Pete L. ClarkDec 16 '13 at 21:50